Find the vector equation of a line passing through the point $(2 \hat{i} - 3 \hat{j} - 5 \hat{k})$ and perpendicular to the plane $r \cdot (6 \hat{i} - 3 \hat{j} + 5 \hat{k}) + 2 = 0$ .
Also, find the point of intersection of this line and the plane.

Question

Find the vector equation of a line passing through the point

(2 \hat{i} - 3 \hat{j} - 5 \hat{k})

and perpendicular to the plane

r \cdot (6 \hat{i} - 3 \hat{j} + 5 \hat{k}) + 2 = 0

.
Also, find the point of intersection of this line and the plane.

Filo · Accepted Answer

Clearly, the required line passes through the point

(2 \hat{i} - 3 \hat{j} - 5 \hat{k})

and is parallel to the normal of the given plane, which is

(6 \hat{i} - 3 \hat{j} + 5 \hat{k})

.

So, the required vector equation is

r = (2 \hat{i} - 3 \hat{j} - 5 \hat{k}) + λ (6 \hat{i} - 3 \hat{j} + 5 \hat{k})

.

The Cartesian equation of the line is

\frac{x - 2}{6} = \frac{y + 3}{- 3} = \frac{z + 5}{5} = k

.

A general point on this line is

P (6 k + 2, - 3 k - 3, 5 k - 5)

.

For same particular value of

k,

let the line cut the plane

6 x - 3 y + 5 z + 2 = 0

.

Then,

6 (6 k + 2) - 3 (- 3 k - 3) + 5 (5 k - 5) + 2 = 0

\Rightarrow (36 k + 9 k + 25 k) = 2 \Rightarrow 70 k = 2

\Rightarrow k = \frac{1}{35}

∴

required point of intersection of the line and plane is

P (\frac{6}{35} + 2, \frac{- 3}{35} - 3, \frac{1}{7} - 5)

, i.e.,

P (\frac{76}{35}, \frac{- 108}{35}, \frac{- 34}{7})

.

Question Text	Find the vector equation of a line passing through the point $(2 \hat{i} - 3 \hat{j} - 5 \hat{k})$ and perpendicular to the plane $r \cdot (6 \hat{i} - 3 \hat{j} + 5 \hat{k}) + 2 = 0$ . Also, find the point of intersection of this line and the plane.
Answer Type	Text solution:1
Upvotes	150

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